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    Hello, could someone please check if I've calculated the inverse of this function properly?

    f( x)=\frac{x^2+1}{x^2-1} domain is (1,3).


    y=\frac{x^2+1}{x^2-1}


    y(x^2-1)=x^2+1


    yx^2-x^2=x^2+1


    yx^2-x^2=x^2+1


    x^2(y-1)=y+1


    x^2=\frac{y+1}{y-1}


    x=\sqrt{\frac{y+1}{y-1}}



    f^{-1}( x)=\sqrt{\frac{x+1}{x-1}}



    Is that correct, and what would be the domain of the inverse function?
    Would it be the same as the original function?
    Thanks.
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    (Original post by hollywoodbudgie)
    Hello, could someone please check if I've calculated the inverse of this function properly?

    f( x)=\frac{x^2+1}{x^2-1} domain is (1,3).


    y=\frac{x^2+1}{x^2-1}


    y(x^2-1)=x^2+1


    yx^2-x^2=x^2+1


    yx^2-x^2=x^2+1


    x^2(y-1)=y+1


    x^2=\frac{y+1}{y-1}


    x=\sqrt{\frac{y+1}{y-1}}



    f^{-1}( x)=\sqrt{\frac{x+1}{x-1}}



    Is that correct, and what would be the domain of the inverse function?
    Would it be the same as the original function?
    Thanks.
    I think your inverse is incorrect. The problem I have with your working is where you went from yx^2-x^2=x^2+1 to x^2(y-1)=y+1. The LHS of that is fine but I'm not sure what you did with the RHS. You've implied that y=x^2, which isn't true.
    The domain of an inverse function is the range of the original function. Writing f(x)=\frac{x^2+1}{x^2-1} = 1+\frac{2}{(x+1)(x-1)} will help with finding the range of f(x).
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    (Original post by Farhan.Hanif93)
    I think your inverse is incorrect. The problem I have with your working is where you went from yx^2-x^2=x^2+1 to x^2(y-1)=y+1. The LHS of that is fine but I'm not sure what you did with the RHS. You've implied that y=x^2, which isn't true.
    The domain of an inverse function is the range of the original function. Writing f(x)=\frac{x^2+1}{x^2-1} = 1+\frac{2}{(x+1)(x-1)} will help with finding the range of f(x).

    The first equation you've mentioned was a typo. It is in fact supposed to be,

    yx^2-y=x^2+1, which comes about through multiplying the bracket.


    So I think that would make it correct.

    Thanks for your tip regarding the domain of the inverse.
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    (Original post by hollywoodbudgie)
    The first equation you've mentioned was a typo. It is in fact supposed to be,

    yx^2-y=x^2+1, which comes about through multiplying the bracket.


    So I think that would make it correct.

    Thanks for your tip regarding the domain of the inverse.
    In which case, yes you're right.
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    (Original post by Farhan.Hanif93)
    In which case, yes you're right.
    Thanks
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    (Original post by Farhan.Hanif93)
    In which case, yes you're right.
    The range of the function is (1,3), which is also its 'domain'.

    The domain of the inverse must therefore be (1,3).

    But that doesn't make sense.

    Shouldn't the domain of the original function be anything other than 1 or -1?
 
 
 
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